• Support PF! Buy your school textbooks, materials and every day products Here!

Trig Identity

  • Thread starter brandon26
  • Start date
  • #1
107
0
Please help me prove this identity:

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.
I cant go any further, please help!:confused:
 

Answers and Replies

  • #2
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
Well, how is cos^2A-sin^2A related to cos2A?
 
  • #3
TD
Homework Helper
1,022
0
And then for the RHS, try changing the tan and the csc to the corresponding sine and/or cosine expressions.
 
  • #4
107
0
arildno said:
Well, how is cos^2A-sin^2A related to cos2A?[/QUOT
one side is equal to the other.
 
  • #5
107
0
I got as far as simplifying the equation to 2cosA / cos2A. What now?
 
  • #6
Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
5,500
7
brandon26 said:
Please help me prove this identity:

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.
So far so good. To finish you'll need the following trig identity:

[tex]\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}[/tex]

You'll want to try to make the LHS of your identity look like that. To do that you'll need to divide the numerator and denominator of your expression by some other expression, and you need to figure out which one that is.

Hint: Look at the [itex]1[/itex] in the denominator above. What would you have to divide [itex]\cos^2(A)[/itex] by to get a [itex]1[/itex]?
 
Last edited:
  • #7
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
The simplest way is now to see:
[tex]\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)[/tex]

I'll leave the last step to you..
 
Last edited:
  • #8
TD
Homework Helper
1,022
0
brandon26 said:
I got as far as simplifying the equation to 2cosA / cos2A. What now?
Once you have this, substitute tan(2a) by sin(2a)/cos(2a) and use the double angle formula on sin(2a). After that, realize that csc(a) = 1/sin(a) and you should be there :smile:
 
  • #9
107
0
arildno said:
The simplest way is now to see:
[tex]\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)[/tex]
I'll leave the last step to you..
tan2A (2cosA/sin2A) = tan2A (2cosA/ (2sinAcosA)) = tan2AcosecA.

Safe boys!:biggrin:
 

Related Threads on Trig Identity

  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
652
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
649
  • Last Post
Replies
7
Views
1K
Top